Is it true that $\limsup_{n\to\infty}(\sup_{x\in B}f_n(x))\leq\sup_{x\in B}(\lim\sup_{n\to\infty}f_n(x))$?

Question

Let $\left(f_n\right)_n$ be a sequence of functions $f_n:A\to [-\infty,\infty]$ and $\varnothing\neq B\subset A$. Is it then true that $\limsup_{n\to\infty}\left(\sup_{x\in B}f_n(x)\right)\leq\sup_{x\in B}(\lim\sup_{n\to\infty}f_n(x))$? This is really a question on whether I understand the definitions of limsup and sup and as of now the answer seems to be no. We can write both expressions as

$$\lim\sup_{n\to\infty}\left(\sup_{x\in B}f_n(x)\right) = \inf_{n\to\infty}\left(\sup_{m\geq n}\left(\sup_{x\in B}f_m(x)\right)\right)\Longleftrightarrow$$

$$\lim\sup_{n\to\infty}\left(\sup_{x\in B}f_n(x)\right) = \inf_{n\to\infty}\left(\sup_{m\geq n,x\in B}f_m(x)\right)$$

and

$$\sup_{x\in B}\left(\lim\sup_{n\to\infty}f_n(x)\right) = \sup_{x\in B}\left(\inf_{n\to\infty}\left(\sup_{m\geq n}f_m(x)\right)\right)$$

If we were to consider the finite version for max and min, I would like to say that the inequality should be reversed for any $x\in A$, maximizer or not, $f_n(x)\leq \max_{x\in A}\{f_n(x)\}$. But I don't know how to properly analyze the limsup in this case.

Answer 1

The reverse inequality does indeed hold. Specifically, for any $x\in B$ and any $n$, $$\sup_{x'\in B}f_n(x')\geqslant f_n(x),$$ so taking "limsups," $$\limsup_{n\to\infty}\sup_{x'\in B}f_n(x')\geqslant \limsup_{n\to\infty}f_n(x).$$ The point $x\in B$ was arbitrary, so $$\limsup_{n\to\infty}\sup_{x\in B}f_n(x)\geqslant \sup_{x\in B}\limsup_{n\to\infty}f_n(x).$$

Answer 2

Try this example: $A = B = \mathbb{N}$, and $$ f_n(x) = \begin{cases} 1, & \text{if } x = n\\ 0, & \text{otherwise.} \end{cases} $$